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In summary, the conversation is about trying to determine the torque needed for an auger that is 6.00 inches in diameter and 44 feet long on a 17% incline. The weight of the product going through the auger is 29lbs per cubic foot and the flow rate is 1200 bushels per hour. The speaker is having difficulty calculating the torque needed and is looking for help. The conversation also references resources and articles that may be useful in making the calculations. However, there is missing information, specifically the pitch of the auger, which is necessary to calculate the torque.
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SevenToFive
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I am trying to figure out the torque needed for an auger, but I seem to run into a wall. The auger is 6.00 inches in diameter and 44 feet long and on a 17% incline. The weight of the product going through the auger is 29lbs per cubic foot. So I calculated the cubic feet of the auger, 3.14 * r^2*H = 3.14*0.5^2*44=34.57 cubic feet. Then 34.57ft^3*29lbs*ft^3 = 1002lbs.
I am having a difficult time figuring out how to calculate the torque needed, even if it is a conservative value to know how large of a gearbox we would need with a 40:1 gear ratio and if we can use a 1HP motor on it. The only other information that I have is that the flow rate is 1200 bushels per hour.
Any help is greatly appreciated.
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haruspex
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From the throughput, the slope and the length you can calculate the power needed. Through what vertical height is the feedstock lifted, what work is done in the process, and at what rate?
For the torque there is not enough info. You need to know the pitch of the auger, i.e. in each turn how far does the feedstock move.
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Asymptotic
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While looking around to see how larger conveyor augers are dimensioned (it appears 1-1/4" ID and 6" pitch is a common size for 6" OD) found an engineering resource http://www.kwsmfg.com/services/screw-conveyor-engineering-guide/kws-calculators.htm# that may be useful. The industry magazine Powder and Bulk Engineering has a variety of articles on the nitty-gritty of conveying, and this one in particular is a good read http://www.powderbulk.com/enews/2013/editorial/story_pdf/pbe_112013RIHF.pdf
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SevenToFive
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Haruspex, thanks for reconfirming that I am missing something to complete the calculations.
Asymptotic, Thanks for the links.
FAQ: Calculate Torque Needed for 6" Auger on 17% Incline
What is torque and why is it important for an auger on an incline?
Torque is a measure of the turning force of an object. In the case of an auger on an incline, torque is important because it determines the amount of power needed to rotate the auger and move material up the incline.
How do I calculate the torque needed for a 6" auger on a 17% incline?
To calculate the torque needed, you will need to know the weight of the material being moved, the angle of the incline, and the diameter of the auger. You can use the formula: torque = (weight * incline * diameter) / (2 * pi).
What units should I use when calculating torque for an auger on an incline?
Torque is typically measured in pound-feet (lb-ft) or Newton-meters (N-m). It is important to use consistent units when calculating torque.
What other factors should I consider when calculating torque for an auger on an incline?
In addition to the weight of the material, incline, and diameter of the auger, you should also consider the efficiency of the auger and the coefficient of friction between the material and the incline surface. These factors can affect the amount of torque needed.
Can I use the same torque calculation for different sizes of augers or inclines?
No, the torque calculation will vary depending on the size of the auger and the angle of the incline. It is important to recalculate the torque for each specific situation to ensure the proper amount of power is being applied.
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